I am currently trying to learn Patterson-Sullivan theory, but I am getting stuck on basic questions about ergodic theory. Here is one of them, given as an exercise in one of the texts I am trying to read. If you need wider context, the text (in French) is here, with the relevant question towards the bottom of page 17 (next-to-last sentence of the long paragraph).

Let ($\Omega$, $\Gamma$, $\mu$) be a dynamical system satisfying the following assumptions:

- $\Omega$ is a Hausdorff, locally compact and $\sigma$-compact topological space;
- $\Gamma$ is a countable group acting on $\Omega$;
- $\mu$ is a Radon measure on $\Omega$ that is quasi-invariant by $\Gamma$ (for every $g \in \Gamma$, $g_* \mu$ is absolutely continuous with respect to $\mu$).

Let us recall some definitions:

- a subset $W \subset \Omega$ is called
*wandering*if, for $\mu$-almost all $w \in W$, the intersection of the orbit $\Gamma w$ with $W$ is finite; - we say that the system is
*completely conservative*if $\Omega$ has no wandering subsets of positive measure; - we say that the system is
*completely dissipative*if $\Omega$ has some wandering subset $W$ such that $\Omega = \bigcup_{g \in \Gamma} g W$.

**Statement to prove: If the system is ergodic and the measure $\mu$ does not have any atoms, then the system is completely conservative.**

Even though it is supposed to be an easy exercise ("on vérifie aisément que..."), I am at a loss. I have found a partial proof, that works only if you can assume that the set $W$ introduced in the proof is closed. Unfortunately, I have no idea whether it is true, in general, that every completely dissipative dynamical system is generated by a closed wandering set. **Can someone give me a proof of this last statement, or find some other way to fix my proof presented below?**

In fact, I have remarkably little intuition about what "completely dissipative" actually means. I am tempted to think that a completely dissipative action should in particular be properly discontinuous, but apparently this is false (though I have never seen an actual counterexample). I would be very happy if I could find, somewhere, an arsenal of examples and counterexamples of completely dissipative and completely conservative systems, that is sufficiently well-stocked to forge an intuition about these properties and to be able to test whether some reasonable-sounding statement is actually true.

**The partial proof**: Let $\Omega = \Omega_C \sqcup \Omega_D$ be the Hopf decomposition of $\Omega$, i.e. a partition of $\Omega$ into two $\Gamma$-invariant subsets such that $\Omega_C$ is completely conservative and $\Omega_D$ is completely dissipative. Since the system is ergodic, they cannot both have positive measure: this means that $\Omega$ is either completely conservative or completely dissipative. By contradiction, assume the latter: let $W \subset \Omega$ be a wandering subset such that
$$\Omega = \bigcup_{g \in \Gamma} g W.$$
Note that the set
$$\{x \in W \mid \Gamma x \cap W \text{ is finite}\} \cap \operatorname{Supp}(\mu)$$
has the same (positive) measure as $W$, hence is nonempty. Let $x_0$ be any point in this set. Let $F := \{ g \in \Gamma \mid g x_0 \in W \}$; this is a finite set by construction.

Since $x_0 \in \operatorname{Supp}(\mu)$, every neighborhood of $x_0$ has positive measure; on the other hand, since the measure is Radon and without atoms, any point has neighborhoods of arbitrarily small measure. Let $U$ be some neighborhood of $x_0$ such that for every $g \in F$, we have $\mu(g U) < \frac{\mu(W)}{\# F}$. We can also assume (**provided that $W$ is closed!**) that for all $g \not\in F$, the image $g U$ is disjoint from $W$. Then consider the set $X := \bigcup_{g \in \Gamma} g U$:

- this set is $\Gamma$-invariant by construction;
- it has positive measure by construction;
- its complement $\Omega \setminus X$ contains in particular $W \setminus X$, the complement of $W \cap X$ in $W$. But by construction, the intersection $W \cap X$ is contained in $\bigcup_{g \in F} g U$, so that we have $$\mu(W \cap X) \leq \sum_{g \in F} \mu(g U) < \mu(W).$$ Hence the complement $W \setminus X$, and hence also the complement $\Omega \setminus X$, has positive measure. But these three properties of the set $X$ contradict ergodicity, QED.